Math  /  Numbers & Operations

Question13. At 5:30 p.m., the temperature in Antarctica is 60F-60^{\circ} \mathrm{F}. At 11:30 p.m., the temperature is 1515^{\circ} cooler. What is the temperature at 11:3011: 30 p.m.? \qquad

Studdy Solution
Perform the subtraction to find the new temperature.
New Temperature=6015=75F \text{New Temperature} = -60 - 15 = -75^{\circ} \mathrm{F}
The temperature at 11:30 p.m. is:
75F\boxed{-75^{\circ} \mathrm{F}}

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